Given that the molecular weight of ethyl alcohol ( $C H_{3} C H_{2} O H$ ) is $46$ , and that of water is $18$ , how many grams of ethyl alcohol must be mixed with $100 m L$ of water for the mole fraction $(X)$ of ethyl alcohol to be $0.2$ ?

74.4 g

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54.4 g

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44.4 g

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64.4 g

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Solution

The correct option is D $64.4 g$
Mole fraction = $\frac{n_{a l c o h o l}}{n_{a l c o h o l} + n_{w a t e r}}$

100 ml $=$ 100 g of water.

Moles of water = $\frac{100}{18} = 5.55 m o l e s$

Mole fraction of ethyl alcohol $= 0.2 = \frac{\frac{w e i g h t}{46}}{\frac{w e i g h t}{46} + 5.55}$

Mass of ethyl alcohol $= 64.4 g$

Hence, option D is correct.

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